A 19.0-micro F capacitor and a 48.0-microF capacitor by being connected across s

Question

A 19.0-micro F capacitor and a 48.0-microF capacitor by being connected across separate 80.0-V batteries. Determine the resulting charge on each capacitor. (Give your answer to at least three significant figures.) 19.0-micro F mC 48.0-micro F mC The capacitors are then disconnected from their batteries and connected to each other, with each negative plate connected to the other positive plate. What is the final charge of each capacitor? 19.0-micro F micro C 48.0-micro F micro C What is the final potential difference across the 48.0-micro F capacitor?

Explanation / Answer

Q=C*V

A)Q1=19*(10^-6)*80=1.52mC

Q2=48*(10^-6)*80=3.84mC

B)here capacitor is connected in series

Ceq=C1*C2/C1+C2=13.612*(10^-6)F

V=1.52+3.84/13.612=39.38V

Q1=19*39.38=7.481*(10^-6)C

Q2=48*39.38=18.901*(10^-6)C

C)V=18.901*(10^-6)/48*(10^-6)=39.37V

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